Rolle's theorem is a special case of the mean value theorem. It is discussed here through examples and questions.
Rolle's Theorem
Rolle's theorem is the result of the mean value theorem where under the conditions:
f(x) be a continuous functions on the interval [a, b] and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a). Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0. In other words, under the above three conditions we can always find a tangent to the curve of f that is horizontal (slope = f'(c) = 0). Rolle's Theorem
If f(x) is
1) a continuous function on the interval [a, b] 2) differentiable on the open interval (a, b) 3) and f(a) = f(b), then there is at least one value c of x in the interval (a, b) such that f '(c) = 0 Example 1The graph of f(x) = - x^{2} + 6x - 6 for 1 ≤ x ≤ 5 is shown below. f(1) = f(5) = - 1 and f is continuous on [1 , 5] and differentiable on (1 , 5) hence, according to Rolle's theorem, there exists at least one value of x = c such that f '(c) = 0.f '(x) = - 2 x + 6 f '(c) = - 2 c + 6 = 0 Solve the above equation to obtain c = 3 Therefore at x = 3 there is a tangent to the graph of f that has a slope equal to zero (horizontal line) as shown in figure 1 below. Example 2The graph of f(x) = sin(x) + 2 for 0 ≤ x ≤ 2π is shown below. f(0) = f(2π) = 2 and f is continuous on [0 , 2π] and differentiable on (0 , 2π) hence, according to Rolle's theorem, there exists at least one value (there may be more than one!) of x = c such that f '(c) = 0.f '(x) = cos(x) f '(c) = cos(c) = 0 The above equation has two solutions on the interval [0 , 2π] c_{1} = π/2 and c_{2} = 3π/2. Therefore both at x = π/2 and x = 3 π/2 there are tangents to the graph that have a slope equal to zero (horizontal line) as shown in figure 2 below. Example 3Function f in figure 3 does not satisfy Rolle's theorem: although it is continuous and f(-1) = f(3), the function is not differentiable at x = 1 and therefore f '(c) = 0 with c in the interval (-1 , 3) is not guaranteed. In fact it is easy to see that there is no horizontal tangent to the graph of f on the interval (-1 , 3).Question 1Which of the functions given below satisfy all three conditions of Rolle's theorem?a) f(x) = cos(x) , for x in [0 , 2π] b) g(x) = |x - 2| , for x in [0 , 4] c) h(x) = 1 / x^{2} , for x in [-1 , 1] d) k(x) = |sin(x)| , for x in [0 , 2π] solution to question 1a)f(0) = 1 and f(2π) = 1 therefore f(0) = f(2π) f is continuous on [0 , 2π] Function f is differentiable in (0 , 2π) Function f satisfies all conditions of Rolle's theorem b) function g has a V-shaped graph with vertex at x = 2 and is therefore not differentiable at x = 2. Function g does not satisfy all conditions of Rolle's theorem c) Function h is undefined at x = 0. Function h does not satisfy all conditions of Rolle's theorem. d) k(x) = |sin(x)| , for x in [0 , 2π] The graph of function k is shown below and it shows that function k is not differentiable at x = π. Function k does not satisfy all conditions of Rolle's theorem. Question 2Check that function f(x) = x^{ 2} - 4 x + 3 on the interval [1 , 3] satisfies all conditions of Rolle's theorem and then find all values of x = c such that f '(c) = 0.solution to question 2f is a polynomial function and is therefore continuous on the interval [1 , 3] and differentiable on the interval (1 , 3). Also f(1) = f(3) = 0 and therefore function f satisfies all three conditions of Rolle's theorem and there is at least one value of x = c such that f '(c) = 0.f '(x) = 2 x - 4 f '(c) = 2 c - 4 = 0 Solve for c to obtain c = 2. The graph below shows f , f ' and the tangent at x = c = 2 is horizontal; and f '(2) = 0 Question 3Check that function g(x) = cos(x) on the interval [- π/2 , 3π/2] satisfies all conditions of Rolle's theorem and then find all values x = c such that g '(c) = 0.solution to question 3Function g is a cosine function and is therefore continuous on the interval [- π/2 , 3π/2] and differentiable on the interval(-π/2 , 3π/2). Also g(- π/2) = g(3π/2) = 0 and therefore function g satisfies all three conditions of Rolle's theorem and there is at least one value of x = c such that f '(c) = 0. g '(x) = - sin(x) g '(c) = - sin(c) = 0 Solve for c to obtain c = nπ , n = 0,~+mn~ 1 , ~+mn~ 2 , ... Solutions on the interval [- π/2 , 3π/2] are c_{1} = 0 and c_{2} = π The graph below shows g , g ' and the tangents at x = c_{1} = 0 and x = c_{2} = π are horizontal; and g '(0) = 0 and g '(π) = 0 |